Problem: Factor completely. $(x^2+x-6)(2x^2+4x)=$
Explanation: The expression is already somewhat factored, because it is given as the product of two factors, $(x^2+x-6)$ and $(2x^2+4x)$. To completely factor the expression, we need to factor each of these factors further. Factoring $(x^2+x-6)$ $x^2+x-6=(x+3)(x-2)$ Factoring $(2x^2+4x)$ These terms have a common factor. The greatest common factor of $2x^2$ and $4x$ is $2x$. Let's factor $2x$ out of $2x^2+4x$ : $\begin{aligned} &\phantom{=}2x^2+4x \\\\ &=2x(x)+2x(2) \\\\ &=2x(x+2) \end{aligned}$ Putting it all together $\begin{aligned} &\phantom{=}{(x^2+x-6)}C{(2x^2+4x)} \\\\ &={(x+3)(x-2)}C{(2x)(x+2)} \\\\ &=2x(x+3)(x-2)(x+2) \end{aligned}$ In conclusion, this is the completely factored expression: $2x(x+3)(x-2)(x+2)$